The first expression, 12C3, is the right answer. There is no limitation saying that the team must be made up of one student from each year level. Furthermore, no distinction has been made among the three players (no assignments of Boards 1, 2, or 3), as the question just asks how many different groups. Since order does not matter, this is a combination. (If order mattered, this would be a permutation).
<span>$7 for the small boxes and $13 for the large</span>
Answer: 4697
Step-by-step explanation:
x + y = 11 ......... i
x² + y² = 325 ........ ii
From equation i
x + y = 11
Therefore, x = 11 - y....... iii
Put equation iii into equation ii
x² + y² = 325
(11 - y)² + y² = 325
(11 - y)(11 - y) + y² = 325
121 - 11y - 11y + y² + y² = 325
121 - 22y + 2y² = 325
121 - 325 - 22y + 2y² = 0
2y² - 22y -204 = 0
2y² - 34y + 12y - 204 = 0
2y(y - 17) + 12(y - 17) = 0
(y - 17) = 0
y = 0 + 17
y = 17
Since x + y = 11
x + 17 = 11
x = 11 - 17
x = -6
Therefore, x³ + y³
= (-6)³ + (17)³
= -216 + 4913
= 4697
Answer:
N = 2
Step-by-step explanation:
n + 5(n-1) = 7
n + 5n - 5 = 7
6n - 5 = 7
6n = 12
n = 2
HOPE THIS HELPS
PLZZ MARK BRAINLIEST!!!
